3 Nov 97Entropy & Free Energy (Ch 20)1 CHEMICAL EQUILIBRIUM Chapter 16 equilibrium vs. completed...

3 Nov 97 Entropy & Free Energy (Ch 20) 1

CHEMICAL EQUILIBRIUMChapter 16

• equilibrium vs. completed reactions• equilibrium constant expressions• Reaction quotient• computing positions of equilibria: examples• Le Chatelier’s principle - effect on equilibria of:

• addition of reactant or product• pressure• temperature

YOU ARE NOT RESPONSIBLE for section 16.7 (relation to kinetics)


For any type of chemical equilibrium of the type

THE EQUILIBRIUM CONSTANTTHE EQUILIBRIUM CONSTANT

K =[C]c [D]d

[A]a [B]b

conc. of products

conc. of reactantsequilibrium constant

If K is known, then we can predict concentrations of products or reactants.

a A + b B c C + d D

the following is a CONSTANT (at a given T) :


All reacting chemical systems can be characterized by their REACTION QUOTIENT, Q.

Q = =0.350.25

= 1.40

Q - the reaction quotient

If Q = K, then system is at equilibrium.

To reach EQUILIBRIUM[Iso] must INCREASE and [n] must DECREASE.

Since K =2.5, system NOT AT EQUIL.

[iso][n]

Q has the same form as K, . . . but uses existing concentrations

n-Butane iso-Butane 0.25 0.35


Q/K

Q

Typical EQUILIBRIUM Calculations

2 general types: a. Given set of concentrations, is system at equilibrium ?

Calculate Q compare to K

1

Q = K

IF:Q > K or Q/K > 1 REACTANTS

Q < K or Q/K < 1 PRODUCTS

Q=K at EQUILIBRIUM


Step 2

Put equilibrium [ ]

into Kc .

Kc = [2x]2

[1.00 - x][1.00 - x] = 55.3

Step 1 Define equilibrium condition in terms of initial condition and a change variable

[H2] [I2] [HI]

At equilibrium 1.00-x 1.00-x 2x

H2(g) + I2(g) 2 HI(g) Kc = 55.3

Step 3. Solve for x. 55.3 = (2x)2/(1-x)2

Square root of both sides & solve gives: x = 0.79

[H2] = [I2] = 1.00 - x = 0.21 M

[HI] = 2x = 1.58 MAt equilibrium


“...if a system at equilibrium is disturbed, the

system tends to shift its equilibrium position

to counter the effect of the disturbance.”

EQUILIBRIUM AND EXTERNAL EFFECTS

• The position of equilibrium is changed when there is a change in:

– pressure– changes in concentration– temperature

• The outcome is governed by

LE CHATELIER’S PRINCIPLE Henri Le Chatelier

1850-1936

- Studied mining engineering

- specialized in glass and ceramics.


• If concentration of one species changes, concentrations of other species CHANGESto keep the value of K the same (at constant T)

• no change in K - only position of equilibrium changes.

Shifts in EQUILIBRIUM : Concentration

ADDING REACTANTS- equilibrium shifts to PRODUCTS

ADDING PRODUCTS- equilibrium shifts to REACTANTS

REMOVING PRODUCTS - often used to DRIVE REACTION TO COMPLETION

- GAS-FORMING; PRECIPITATION


Solution

A. Calculate Q with extra 1.50 M n-butane.

INITIALLY: [n] = 0.50 M [iso] = 1.25 M

CHANGE: ADD +1.50 M n-butane What happens ?What happens ?

Q < K . Therefore, reaction shifts to PRODUCT

Q = [iso] / [n] = 1.25 / (0.50 + 1.50) = 0.63

n-Butane Isobutane

Effect of changed [ ] on an equilibrium

16_butane.mov(16m13an1.mov)

K = [iso]

[n] = 2.5


Solution

B. Solve for NEW EQUILIBRIUM - set up concentration table [n-butane] [isobutane]Initial 0.50 + 1.50 1.25Change - x + xEquilibrium 2.00 - x 1.25 + x

Butane/Isobutane

K = 2.50 = [isobutane]

[butane]

1.25 + x2.00 - x

x = 1.07 M. At new equilibrium position, [n-butane] = 0.93 M [isobutane] = 2.32 M.

Equilibrium has shifted toward isobutane.

AB


Effect of Pressure (gas equilibrium)

Increase P in the system by reducing the volume.

Kc = [NO2 ]2

[N2O4 ] = 0.0059 at 298 K

NN22OO44(g) (g) 2 NO 2 NO22(g)(g)

16_NO2.mov(16m14an1.mov)

Increasing P shifts equilibrium to side with fewer molecules (to try to reduce P). Here, reaction shifts LEFT PN2O4 increases

See Ass#2 - question #6

PNO2 decreases



• Temperature change change in K• Consider the fizz in a soft drink

CO2(g) + H2O(liq) CO2(aq) + heat

• Increase TEquilibrium shifts left: [CO2(g)] [CO2 (aq)] K decreases as T goes up.

•Decrease T[CO2 (aq)] increases and [CO2(g)] decreases.K increases as T goes down

Kc = [CO2(aq)]/[CO2(g)]HIGHER T

LOWER T

• Change T: New equilib. position? New value of K?


Temperature Effects on Chemical Equilibrium

Kc = 0.00077 at 273 K

Kc = 0.00590 at 298 KKc

[NO2 ]2

[N2O4 ]

N2O4 + heat 2 NO2 (colorless) (brown)

Horxn = + 57.2 kJ

Increasing T changes K so as to shift equilibrium in ENDOTHERMIC direction

16_NO2RX.mov(16m14an1.mov)



• Add catalyst ---> no change in K• A catalyst only affects the RATE of

approach to equilibrium.

Catalytic exhaust system


CHEMICAL EQUILIBRIUMChapter 16

• equilibrium vs. completed reactions• equilibrium constant expressions• Reaction quotient• computing positions of equilibria: examples• Le Chatelier’s principle - effect on equilibria of:

• addition of reactant or product• pressure• temperature

YOU ARE NOT RESPONSIBLE for section 16.7 (relation to kinetics)


Entropy and Free Energy (Kotz Ch 20)

• Spontaneous vs. non-spontaneous• thermodynamics vs. kinetics• entropy = randomness (So)• Gibbs free energy (Go)• Go for reactions - predicting spontaneous direction

• Grxn versus Gorxn

• predicting equilibrium constants from Gorxn


Entropy and Free Energy ( Kotz Ch 20 )

• How can we predict if a reaction can occur, given enough time?

• Note: Thermodynamics DOES NOT say how quickly (or slowly) a reaction will occur.

• To predict if a reaction can occur at a reasonable rate, one needs to consider:

• some processes are spontaneous; others never occur. WHY ?

THERMODYNAMICS

KINETICS

9-paper.mov20m02vd1.mov


Thermodynamics

• state of a chemical system (P, T, composition)

• From a given state, would a chemical reaction decrease the energy of the system?

• If yes, system is favored to react — a product-favored system which will have a spontaneous reaction.

• Most product-favored reactions are exothermic.

• Spontaneous does not imply anything about time for reaction to occur. (kinetics)


Thermodynamics versus Kinetics

• Diamond to Graphite– spontaneous from thermodynamics

– but not kinetically favored.

• Paper burns.- product - favored reaction.

- Also kinetically favored once reaction is begun.


Product-Favored Reactions

E.g. thermite reaction

Fe2O3(s) + 2 Al(s)

2 Fe(s) + Al2O3(s)

H = - 848 kJ

In general, product-favored reactions are exothermic.


Non-exothermic spontaneous reactions

But many spontaneous reactions or processes are endothermic . . .

NH4NO3(s) + heat NH4+ (aq) + NO3

- (aq)Hsol = +25.7 kJ/mol

or have H = 0 . . .


Entropy, SEntropy, SOne property common to

product-favored processes is that the final state is more DISORDERED or RANDOM than the original.

Spontaneity is related to an increase in randomness.

The thermodynamic property related to randomness is ENTROPY, S. Reaction of K

with water


PROBABILITY - predictor of most stable state

WHY DO PROCESSES with H = 0 occur ?

Consider expansion of gases to equal pressure:

This is spontaneous because the final state,with equal # molecules in each flask, is much more probable than the initial state,with all molecules in flask 1, none in flask 2

SYSTEM CHANGES to state of HIGHER PROBABILITYTHIS IS USUALLY the more RANDOM state.


Gas expansion - spontaneity from greater probability

Consider distribution of 4 molecules in 2 flasks

P1 < P2 P1 > P2P1 = P2

With more molecules (>1020) P1=P2 is most probable by far


WHAT about EXOTHERMIC REACTIONS ?

Consider 2 H2 (g) + O2 (g) 2 H2O (l)• HIGHLY EXOTHERMIC• final state (liquid) is MUCH MORE ORDERED (less random arrangement) than initial state (2 gases) • BUT PROBABILITY of final state is higher when one considers change in the surroundings. WHY ?• Heat evolved increases motion of molecules in the surroundings

>Increased disorder of

surroundings decreased disorder of

system

MUST consider change in disorder in BOTH SYSTEM and SURROUNDINGS to predict DIRECTION of SPONTANEITY


Directionality of Reactions

How probable is it that reactant molecules will react?

PROBABILITY suggests that a product-favored reaction will result in the dispersal of energy or dispersal of matter or both.


Probability suggests that a product-favored reaction will result in the dispersal of energy or of matter or both.

Directionality of Reactions

Energy DispersalMatter Dispersal9_gasmix.mov20m03an1.mov

9_exorxn.mov20m03an2.mov


Directionality of ReactionsEnergy Dispersal

• Exothermic reactions involve a release of stored chemical potential energy to the surroundings.

• The stored potential energy starts out in a few molecules but is finally dispersed over a great many molecules.

• The final state—with energy dispersed—is more probable and makes a reaction product-favored.


Standard Entropies, So

• Every substance at a given temperature and in a specific phase has a well-defined Entropy• At 298o the entropy of a substance is called

So - with UNITS of J.K-1.mol-1

• The larger the value of So, the greater the degree of disorder or randomness

e.g. So (in J.K-1mol-1) : Br2 (liq) = 152.2

Br2 (gas) = 245.5

For any process: So = So(final) - So(initial)

So(vap., Br2) = (245.5-152.2) = 93.3 J.K-1mol-1


S (gases) > S (liquids) > S (solids)

So (J/K•mol)

H2O(gas) 188.8

H2O(liq) 69.9

H2O (s) 47.9

Ice Water

Vapour

Entropy, S

3 Nov 97Entropy & Free Energy (Ch 20)1 CHEMICAL EQUILIBRIUM Chapter 16 equilibrium vs. completed...

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